Introduction

In this blog post, we will explore how to solve the LeetCode problem "1869" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a binary string s, return true if the longest contiguous segment of 1's is strictly longer than the longest contiguous segment of 0's in s, or return false otherwise. For example, in s = "110100010" the longest continuous segment of 1s has length 2, and the longest continuous segment of 0s has length 3. Note that if there are no 0's, then the longest continuous segment of 0's is considered to have a length 0. The same applies if there is no 1's. Example 1: Input: s = "1101" Output: true Explanation: The longest contiguous segment of 1s has length 2: "1101" The longest contiguous segment of 0s has length 1: "1101" The segment of 1s is longer, so return true. Example 2: Input: s = "111000" Output: false Explanation: The longest contiguous segment of 1s has length 3: "111000" The longest contiguous segment of 0s has length 3: "111000" The segment of 1s is not longer, so return false. Example 3: Input: s = "110100010" Output: false Explanation: The longest contiguous segment of 1s has length 2: "110100010" The longest contiguous segment of 0s has length 3: "110100010" The segment of 1s is not longer, so return false. Constraints: 1 <= s.length <= 100 s[i] is either '0' or '1'.

Explanation

Here's the breakdown of the solution:

• Iterate and Track: The core idea is to traverse the string character by character, keeping track of the current consecutive lengths of both '1's and '0's.
• Maintain Maximums: Simultaneously, we update the maximum lengths encountered for both '1's and '0's.

• Compare and Return: Finally, we compare the maximum lengths of '1's and '0's and return True if the '1's are longer, and False otherwise.

• Runtime Complexity: O(n), where n is the length of the string. Storage Complexity: O(1) - constant extra space.

Code

    def check_zero_ones(s: str) -> bool:
    """
    Given a binary string s, return true if the longest contiguous segment of 1's
    is strictly longer than the longest contiguous segment of 0's in s, or return false otherwise.

    For example, in s = "110100010" the longest continuous segment of 1s has length 2,
    and the longest continuous segment of 0s has length 3.

    Note that if there are no 0's, then the longest continuous segment of 0's is considered to have a length 0.
    The same applies if there is no 1's.

    Example 1:
    Input: s = "1101"
    Output: true
    Explanation:
    The longest contiguous segment of 1s has length 2: "1101"
    The longest contiguous segment of 0s has length 1: "1101"
    The segment of 1s is longer, so return true.

    Example 2:
    Input: s = "111000"
    Output: false
    Explanation:
    The longest contiguous segment of 1s has length 3: "111000"
    The longest contiguous segment of 0s has length 3: "111000"
    The segment of 1s is not longer, so return false.

    Example 3:
    Input: s = "110100010"
    Output: false
    Explanation:
    The longest contiguous segment of 1s has length 2: "110100010"
    The longest contiguous segment of 0s has length 3: "110100010"
    The segment of 1s is not longer, so return false.

    Constraints:
    1 <= s.length <= 100
    s[i] is either '0' or '1'.
    """

    max_ones = 0
    max_zeros = 0
    current_ones = 0
    current_zeros = 0

    for char in s:
        if char == '1':
            current_ones += 1
            current_zeros = 0  # Reset zero count
        else:
            current_zeros += 1
            current_ones = 0  # Reset one count

        max_ones = max(max_ones, current_ones)
        max_zeros = max(max_zeros, current_zeros)

    return max_ones > max_zeros